Asked by Dante
                a rancher has 1200 fet of fencing to enclose two adjacent cattle pens. What dimensions should be used so that the enclosed area is maximized?
            
            
        Answers
                    Answered by
            Reiny
            
    You don't describe the diagram, but I am familiar with this type of question.
Assuming there is a common width between the two fields
let the width be x ft
let the length of the combined field be y
so
3x + 2y = 1200
y = 600 - (3/2)x
area = xy
= x(600 - (3/2)x)
= 600x - (3/2)x^2
d(area)/dx = 600 - 3x
= 0 for a max of area
3x = 600
x = 200
y = 600-(3/2)(200) = 300
state the conclusion
    
Assuming there is a common width between the two fields
let the width be x ft
let the length of the combined field be y
so
3x + 2y = 1200
y = 600 - (3/2)x
area = xy
= x(600 - (3/2)x)
= 600x - (3/2)x^2
d(area)/dx = 600 - 3x
= 0 for a max of area
3x = 600
x = 200
y = 600-(3/2)(200) = 300
state the conclusion
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.